Answer: 21.91 s
Explanation:
Given that,
Maximum height of the car, h = 48 ft
Acceleration of the elevator, a = 0.6 ft/s²
Deceleration of the elevator, -a = 0.3 ft/s²
Maximum speed of the elevator, v = 8 ft/s
Initial speed of the elevator, u = 0
If when the elevator accelerate from 0 to maximum velocity, v.
Let s be the vertical distance traveled during acceleration.
v² = u² - 2as
s = (v² - u²) / 2a
s = (8² - 0) / 2*0.6
s = 64 / 1.2
s = 53.33 ft
If when the elevator decelerates from maximum velocity, v to zero.
Let S be the vertical distance traveled during deceleration
u² = v² + 2aS
S = (u² - v²) / 2a
S = (0 - 8²) / 2 * 0.3
S = -64 / 0.6
S = 106.67 ft
Since he sum of s and S (i.e s + S) is greater than 48 ft, then the elevator will switch from acceleration to deceleration
without reaching the maximum velocity. Below, the switching point is labeled y.
v² = u² + 2ay
y = v²/2a
Inserting this into the earlier deceleration equation, we have
-v²/2 = d * [48 - (v²/2a)], where
d = deceleration
a = acceleration
Therefore, v = [4.√6. a √-(a.b/a)] / b
Where b = acceleration - deceleration
v = 4.382 ft/s
Using this newly found v, we proceed to find our s
s = (u² + v²)/2a
s = 19.2 / 1.2
s = 16 ft
The transport times for each segment are found from
v = u + a*t, thus upward t1
4.382 = 0 + 0.6 * t
t = 4.382/0.6
t = 7.303 s
Also,
4.382 = 0 + 0.3 * T
T = 4.382/0.3
T = 14.607 s
The total travel time is then t + T =
7.303 + 14.607
Total time of travel is 21.91 s
